## UVA Problem 12015 – Google is Feeling Lucky Solution

UVA Problem 12015 – Google is Feeling Lucky Solution:

Solving Technique:

This is a sorting problem. In this problem we need to find biggest relevance number ( the second input / parameter). Then print the strings associated with that number. Multiple numbers can have equal relevance in which case print all the matching strings.

#### Ideas:

This can be solved efficiently by sorting inputs and based on that only print the strings / web addresses with biggest relevance number. Then if a lower relevance number is found stop processing. Since it was sorted ( ascending – start from last index, descending – start from index 0 )  if any lower value is found, further processing won’t yield any large relevance value.

Another method is using priority queue and store elements in key value pair. Then pop the biggest values and print their string.

Here my implementation is simple get the max value. Then brute force search all inputs for that value. If any value match the biggest relevance value print the string.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

Input:

```2
www.alibaba.com 10
www.taobao.com 5
www.good.com 7
www.fudan.edu.cn 8
www.university.edu.cn 9
acm.university.edu.cn 10
www.alibaba.com 11
www.taobao.com 5
www.good.com 7
www.fudan.edu.cn 8
acm.university.edu.cn 9
acm.university.edu.cn 10```

Output:

```Case #1:
www.alibaba.com
acm.university.edu.cn
Case #2:
www.alibaba.com```

### Code:

```/**
* @author  Quickgrid ( Asif Ahmed )
* Problem: UVA 12015 Google is feeling lucky
*/

#include<stdio.h>

static char s[10][100];
static unsigned r[10];

static unsigned c, n, max, i;

int main(){

scanf("%u", &n);

while(n--){
scanf("%s %u", &s[0], &r[0]);
max = r[0];

for(i = 1; i < 10; ++i){
scanf("%s %u", &s[i], &r[i]);
if(r[i] > max)
max = r[i];
}

printf("Case #%u:\n", ++c);
for(i = 0; i < 10; ++i){
if(r[i] == max)
printf("%s\n", s[i]);
}
}
return 0;
}
```